Question

If $\alpha$ and $\beta$ are angles in the first quadrant, $\tan \alpha =\frac{1}{7},\sin \beta =\frac{1}{\sqrt{10}}$ then using the formula $\sin (A+B)=\sin A\cos B+\cos A\sin B$ the value of $(\alpha +2\beta )$ is

• A. $0^{\circ }$
• B. ${45}^{\circ }$
• C. ${60}^{\circ }$
• D. ${90}^{\circ }$

Answer 1

The correct option is B ${45}^{\circ }$

Since $\tan \alpha =\frac{1}{7}$

$\therefore \sin \alpha =\frac{1}{5\sqrt{2}}and\cos \alpha =\frac{7}{5\sqrt{2}}$

Since $\sin \beta =\frac{1}{\sqrt{10}}$

$\therefore \cos \beta =\frac{3}{\sqrt{10}}$

$\therefore \sin (\alpha +2\beta )=\sin \alpha \cos \alpha 2\beta +\cos \alpha \sin 2\beta$

$=\sin \alpha (2{\cos }^2\beta -1)+\cos \alpha .2\sin \beta \beta \cos \beta$

$=\frac{1}{5\sqrt{2}}(2\times \frac{9}{10}-1)+\frac{7}{5\sqrt{2}}\times 2\times \frac{1}{\sqrt{10}}\times \frac{3}{\sqrt{10}}$

$=\frac{1}{5\sqrt{2}}\times \frac{4}{5}+\frac{21}{25\sqrt{2}}=\frac{4}{25\sqrt{2}}+\frac{21}{25\sqrt{2}}$

$=\frac{25}{25\sqrt{2}}=\frac{1}{\sqrt{2}}=\sin {45}^{\circ }$

or $(\alpha +2\beta )={45}^{\circ }$